Q: Solve given equation log 2 3 plus log 2 a log 2 20?

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it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.

log(f) + log(0.1) = 6 So log(f*0.1) = 6 so f*0.1 = 106 so f = 107

It is my first answer. Is the problem to solve A=B^X ? where A and B are positive integers and X the power exponent of B The given equation can be rewritten in a logarithm form. Log A = X * Log B solving for a unique X X = Log A / Log B The result: Any positive integer A can be rewritten as a positive integer B to the distinct power X. Where X is Log A divided by Log B A = B ^(Log A / Log B) I think, this is the solution. Roger Verbeeck

I will assume that the "2" after each log is a subscript (indicating log to the base 2). Basically, you must use the well-known logarithmic identities, a log b = log ba, and log a + log b = log ab. 2 log2x + log21 = log24 log2x2 + log21 = log24 log2x2(1) = log24 log2x2 = log24 Take antilogs on both sides: x2 = 4 In this last equation, x is either 2 or -2. However, negative numbers are not appropriate for the original equation (assuming real numbers), so the only solution is 2. For safety, this should be checked in the original equation; I'll leave that part to you.

Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.

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it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.

The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)

9x = 27 log(9) + log(x) = log(27) log(x) = log(27) - log(9) log(x) = log(27/9) 10log(x) = 10log(27/9) x = 27/9 x = 3 This strikes us as the method by which the federal government might solve the given equation ... after appointing commissions to study the environmental impact and recommend a method of solution, of course.

You calculate a log, you do not solve a log!

log1 + log2 + log3 = log(1*2*3) = log6

You can calculate the pKa value by using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid. Rearranging the equation, you can solve for pKa by taking the antilog of both sides after isolating pKa.

log(f) + log(0.1) = 6 So log(f*0.1) = 6 so f*0.1 = 106 so f = 107

It is my first answer. Is the problem to solve A=B^X ? where A and B are positive integers and X the power exponent of B The given equation can be rewritten in a logarithm form. Log A = X * Log B solving for a unique X X = Log A / Log B The result: Any positive integer A can be rewritten as a positive integer B to the distinct power X. Where X is Log A divided by Log B A = B ^(Log A / Log B) I think, this is the solution. Roger Verbeeck

I will assume that the "2" after each log is a subscript (indicating log to the base 2). Basically, you must use the well-known logarithmic identities, a log b = log ba, and log a + log b = log ab. 2 log2x + log21 = log24 log2x2 + log21 = log24 log2x2(1) = log24 log2x2 = log24 Take antilogs on both sides: x2 = 4 In this last equation, x is either 2 or -2. However, negative numbers are not appropriate for the original equation (assuming real numbers), so the only solution is 2. For safety, this should be checked in the original equation; I'll leave that part to you.

Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.

I assume you are asking how to solve a logarithmic equation. Well let's quickly review what the log function is: for the equation log(x)=y, we are saying that 10^y=x. So once you have isolated the logarithm, take the value of the base, raise it to the nth power (when 'n' is the value that the function is equal to) and set that equal to the value inside of the log.

2 log(x) + 3 log(x) = 105 log(x) = 10log(x) = 10/5 = 210log(x) = (10)2x = 100